Separate (9x^2 + 8x + 10)/(x^2 + 1)(x + 2) into partial fractions.

First, we find the form of the two fractions we're going to get. As one denominator has a power of 2, and the other a power of 1, our answer will be of the form: [(Ax+B)/(x²+1)] + [C/(x+2)]. If we make this our Right Hand Side, and make the question's equation our Left Hand Side, and then multiply by our LHS denominator, we get 9x²+8x+10=(Ax+B)(x+2)+C(x²+1). If we then set x = -2, to eliminate our Ax+B term, so that we can solve for C, our equation becomes 9(-2)²+8(-2)+10=C[(-2)²+1], Simplifying, we get 30=5C, and solve C=6. We then put this back into the initial equation we made and move the C fraction to the other side, we get (Ax+B)/(x²+1) = [(9x²+8x+10)/(x+2)(x²+1)] - [6/(x+2)]. We then make our RHS into one single fraction, by multiplying the latter fraction, top and bottom, by (x²+1). If we then simplify the numerator of our new RHS, we get (3x²+8x+4)/(x+2)(x²+1). We can factorise the numerator to give us (3x+2)(x+2), and find that the x+2 cancels out with the denominator. Now, our whole equation becomes (Ax+B)/(x²+1) = (3x+2)/(x²+1). Equating our coefficients of x, we find that A = 3, B = 2. So, our final solution is: (9x²+8x+10)/(x+2)(x²+1) = [(3x+2)/(x²+1)] + [6/(x+2)].